If the solution curve of the differential equation (y-2 log_ex)dx+(x log_e x^2)dy=0,x>1 passes through the points (e, 4 3) and (e^4,α ), then α is…

Given,

$$(y-2{\mathrm{log}}_{e}x)dx+(x{\mathrm{log}}_{e}{x}^{2})dy=0,x>1$$(y−2loge​x)dx+(xloge​x2)dy=0,x>1

$$\Rightarrow 2x\mathrm{ln}x\frac{dy}{dx}+y=2\mathrm{ln}x{\text{where}{\mathrm{log}}_{e}x=\mathrm{ln}x}$$⇒2xlnxdxdy​+y=2lnxwhereloge​x=lnx

$$\Rightarrow \frac{dy}{dx}+\frac{y}{2x\mathrm{ln}x}=\frac{1}{x}$$⇒dxdy​+2xlnxy​=x1​

Which is linear differential equation,

So, $\displaystyle I.F={e}^{\int \frac{1}{2x\mathrm{ln}x}}=\sqrt{\mathrm{ln}x}$I.F=e∫2xlnx1​=lnx​

Now, solution of the equation is given by,

$$y\cdot \sqrt{\mathrm{ln}x}=\int \frac{\sqrt{\mathrm{ln}x}}{x}dx$$y⋅lnx​=∫xlnx​​dx

$$\Rightarrow y\cdot \sqrt{\mathrm{ln}x}=\frac{2}{3}{(\mathrm{ln}x)}^{\frac{3}{2}}+C...(i)$$⇒y⋅lnx​=32​(lnx)23​+C...(i)

Given, $eq.(i)$eq.(i) passes through point $(e,\frac{4}{3})$(e,34​)

So, $C=\frac{2}{3}$C=32​

Hence, solution of differential equation will be,

$$y\sqrt{\mathrm{ln}x}=\frac{2}{3}{(\mathrm{ln}x)}^{\frac{3}{2}}+\frac{2}{3}$$ylnx​=32​(lnx)23​+32​

Also given, above equation passes through point $({e}^{4},\alpha )$(e4,α)

$$\alpha \sqrt{\mathrm{ln}{e}^{4}}=\frac{2}{3}{(\mathrm{ln}{e}^{4})}^{\frac{3}{2}}+\frac{2}{3}$$αlne4​=32​(lne4)23​+32​

$$\Rightarrow 2\alpha =\frac{2}{3}{(4)}^{\frac{3}{2}}+\frac{2}{3}$$⇒2α=32​(4)23​+32​

$$\Rightarrow \alpha =\frac{1}{3}\times 8+\frac{1}{3}$$⇒α=31​×8+31​

$$\Rightarrow \alpha =3$$⇒α=3